Question

Pure $P{Cl}_5$ is introduced into an evacuated chamber and comes to equilibrium at ${247}^{o}C$ and $2.0$ $atm$. The equilibrium gaseous mixture contains $40\%$ chlorine by volume.
Calculate $K_p$ at ${247}^{o}C$ for the reaction
$P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$

• A. $0.625$ $atm$
• B. $4$ $atm$
• C. $1.6$ $atm$
• D. $Noneofthese$

Answer 1

Answer: (C)

$1.6$ $atm$

The gaseous mixture contains $40\%C{l}_2$ and $40\%PC{l}_3$, since they are produced in $1:1$ mole ratio. The $PC{l}_5\%$ is $20$.
For ideal gases mole $\%=$ volume $\%$
$P_{C{l}_2}=P_{C{l}_3}$
$\Rightarrow 2\times 0.40=0.80atm$
$P_{C{l}_5}=2\times 0.2=0.40atm$
$\therefore K_P=\frac{P_{PC{l}_3}.P_{C{l}_2}}{P_{PC{l}_5}}$
$=\frac{0.80\times 0.80}{0.40}$
$=1.6atm$